Limit of Sequence

**utopiaNow** · October 25th 2009, 01:58 PM

Prove that if $a,b$ are positive real numbers then $\displaystyle\lim_{n\rightarrow \infty} (a^n +b^n)^{1/n} = \max{(a,b)}$ .

My attempt is to say that without loss of generality let $a > b$ , then for sufficiently large $n$ we will have $a^n + b^n \approx a^n$ hence $\displaystyle\lim_{n\rightarrow \infty} (a^n +b^n)^{1/n} = \displaystyle\lim_{n\rightarrow \infty}(a^n)^{1/n} = a$ . However this step "for sufficiently large $n$ , we will have $a^n + b^n \approx a^n$ " does not seem rigorous to me, and I can't think of a way of proving this result more rigorously.

Any suggestions would be greatly appreciated.

**tonio** · October 25th 2009, 02:18 PM

Originally Posted by utopiaNow

Prove that if $a,b$ are positive real numbers then $\displaystyle\lim_{n\rightarrow \infty} (a^n +b^n)^{1/n} = \max{(a,b)}$ .

My attempt is to say that without loss of generality let $a > b$ , then for sufficiently large $n$ we will have $a^n + b^n \approx a^n$ hence $\displaystyle\lim_{n\rightarrow \infty} (a^n +b^n)^{1/n} = \displaystyle\lim_{n\rightarrow \infty}(a^n)^{1/n} = a$ . However this step "for sufficiently large $n$ , we will have $a^n + b^n \approx a^n$ " does not seem rigorous to me, and I can't think of a way of proving this result more rigorously.

Any suggestions would be greatly appreciated.

As you did, assume wrg that a>b and then "take a out of the parentheses" :

$\sqrt[n]{a^n+b^n}=a\sqrt[n]{1+\left(\frac{b}{a}\right)^n}$ and remember what happens when we have a geometric sequence with quotient q s.t. $|q|<1$

Tonio

Thread: Limit of Sequence - max(a,b)

LinkBack

Thread Tools

Display

Limit of Sequence - max(a,b)

Similar Math Help Forum Discussions

limit of function of a sequence equals the function of the limit of that sequence?

limit of a sequence

Limit of a Sequence

Limit of sequence

Search Tags